Question

The magnetic field at the centre of a current carrying circular coil of area '$A$' is '$B$'. The magnetic moment of the coil is ($\mu_0 = $ permeability of free space)

• A. $\frac{2BA^{3/2}}{\mu_0\pi^{1/2}}$
• B. $\frac{BA^{3/2}}{\mu_0\pi}$
• C. $\frac{\mu_0\pi^{1/2} BA^2}{2}$
• D. $\frac{2BA^2}{\mu_0\pi}$

Hint:

To easily check the dimensional consistency of these types of options, remember that $M$ requires an extra $A$ ($m^2$) on top of whatever dimensions create $I$ from $B$. The $A^{3/2}$ perfectly balances the $A^{1/2}$ in the denominator from $R$.

Answer 1

The Correct Option is A

Step 1: Understand the question.
A circular coil of area $A$ carries a current that makes a magnetic field $B$ at its centre. We must find the coil's magnetic moment, written using $B$, $A$, and constants.

Step 2: Write the three basic facts.
The field at the centre of a circular loop is $B = \dfrac{\mu_0 I}{2R}$. The area is $A = \pi R^2$. The magnetic moment is $M = I A$.

Step 3: Get the radius from the area.
From $A = \pi R^2$:
\[ R = \sqrt{\frac{A}{\pi}} = \frac{A^{1/2}}{\pi^{1/2}} \]

Step 4: Find the current.
Put this $R$ into the field formula and solve for $I$:
\[ B = \frac{\mu_0 I}{2R} = \frac{\mu_0 I\,\pi^{1/2}}{2A^{1/2}} \]
So:
\[ I = \frac{2B A^{1/2}}{\mu_0\,\pi^{1/2}} \]

Step 5: Find the magnetic moment.
Multiply the current by the area:
\[ M = I A = \frac{2B A^{1/2}}{\mu_0\,\pi^{1/2}}\times A \]

Step 6: Combine the area powers.
Since $A^{1/2}\times A = A^{3/2}$:
\[ M = \frac{2B A^{3/2}}{\mu_0\,\pi^{1/2}} \]
This matches option (1).
\[ \boxed{M = \frac{2BA^{3/2}}{\mu_0\pi^{1/2}}} \]